\newproblem{lay:5_3_28}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.3.28}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that if $A$ has $n$ linearly independent eigenvectors, then so does $A^T$. [\textit{Hint}: Use the Diagonalization Theorem.]
}{
   % Solution
	The Diagonalization Theorem states that $A$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors, and that in that
	case, $A$ can be expressed as
	\begin{center}
		$A=PDP^{-1}$,
	\end{center}
	where the columns of $P$ are the $n$ linearly independent eigenvectors. Taking the transpose in both sides, we have
	\begin{center}
		$A^T=(PDP^{-1})^T=(P^{-1})^TD^TP^T=(P^T)^{-1}D^TP^T$
	\end{center}
	So, $A^T$ is also diagonalizable and, by the Diagonalization Theorem again, it must have $n$ linearly independent eigenvectors.
}
\useproblem{lay:5_3_28}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
